Application of Probability to Mechanical Design
121
Then
S(50%) -- x -- 0.7229780 ÷
The parameters used
7=3314psi
0=4878.47 fl=1.1299
S(50%) = 6,841 psi
Now 3 is Eq. 2.133 with/~=0.7015 in Eq. (2.144)
.06066 K~=PL
b 3
2,904
(2.149)
(2.15o)
Substituting into in Eq. (2.148) with Eq. (2.149), (2.150).
N - 6,841 ps~x
2,904
N = 2.36
(2.151)
These Weibull cross section dimensions and safety factors are compared in
Table 2.12 with the Gaussian solutions. The Gaussian computer result is
for iterations starting from the low side.
Gaussian
/~ = 0.875 in
(2.152)
bmax = 1.02576/~
bmax = 0.8975
(2.153)
1.750"
(2.154)
hmax -- 2bmax -- 1.795 in
The factor of safety N is
N==
S
(2.155)
(2.156)
Table 2.12 Cross sectional dimensions Examples 2.21 for Weibull and
Gaussian Monte Carlo simulation
Solution
~ (in)
bmax (in)
~ (in)
hma× (in)
N
Weibull
0.7015
0.7196
1.4030
1.4391
2.36
Gaussian
0.875
0.8975
1.750
1.795
3.86